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Homework Help: How do I rewrite sin(x^1/2)?

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    I would like to know how to rewrite sin[x^(1/2)]? I have to do a laplace transform of a certain equation but I need to get it into a more usable form...


    2. Relevant equations



    3. The attempt at a solution

    I was thinking of using the binomial theorem but 1/2 is neither a even or odd number so I can't use it?
     
  2. jcsd
  3. May 8, 2010 #2

    rock.freak667

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    I don't think there is a way to get in a more usable form. Was your equation to solve directly involving sin(√x)?
     
  4. May 8, 2010 #3
    essentially, yes, where x = at. and

    f(t) = sin(√at)

    I didn't know how to solve it because there was a t^(1/2) term X_X. Do you have any ideas?

    Can I just use the formula L(f) = F(s) = [tex]\int[/tex]f(t)e-st dt where the integration runs from 0 to infinity, even though I have a t^(1/2) term?
     
    Last edited: May 8, 2010
  5. May 8, 2010 #4

    gabbagabbahey

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    Of course. That is the definition of the Laplace transform after all (you are missing your integration limits though)
     
  6. May 9, 2010 #5
    can you have a fourier transform with a complex number in it? The answer I got has a complex number in it...
     
  7. May 9, 2010 #6

    gabbagabbahey

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    If [itex]a[/itex] is real, your answer should be too. When you take the Laplace transform of a real-valued function, you get a real-valued function.

    Maybe you should show us your calculation...
     
  8. May 9, 2010 #7
    You could try 'u' substitution
     
  9. May 9, 2010 #8
    well I'm not sure how to do all the symbols and stuff very well, but basically, I did

    [tex]\int[/tex]sin[(at)1/2]e-stdt

    substituted 1/(2i) * ( exp[i(at)1/2] - exp[-i(at)1/2] ) for the sine term, integrated with limits between 0 and infinity, and eventually got

    -i / s

    as my answer, where i is an imaginary number.
     
  10. May 9, 2010 #9

    gabbagabbahey

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    See my sig :wink:

    Your method looks fine, but your result doesn't....show the rest of your calculations
     
  11. May 9, 2010 #10
    Okay!

    1/(2i) * [ [tex]\int[/tex] exp[i(at)1/2] - st] dt - [tex]\int[/tex] exp[-i(at)1/2] - st] dt ] =

    1/(2i) * [ 1 / [ i(a / 2(at)1/2) - s] - 1 / [ -i(a / 2(at)1/2) - s] ]

    *note at this point I plugged in 0 for t for the lower limit and infinity for the upper limit of t from integration*
    = 1/(2i) * [1/s + 1/s]
    = 1/(2i) * [2/s]
    = 1/is = -i/s
     
  12. May 9, 2010 #11

    gabbagabbahey

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    You seem to be claiming that

    [tex]\int e^{h(t)}dt=\frac{e^{h(t)}}{h'(t)}[/tex]

    Is that really true? (Use the product rule to differentiate [tex]\frac{e^{h(t)}}{h'(t)}[/tex] and see if you get [itex]e^{h(t)}[/tex])

    Instead, start with the substitution [itex]u=\sqrt{at}[/itex] and then complete the square on the exponent...
     
  13. May 9, 2010 #12
    I did the substitution and now I have an exponent equal to

    exp(iu - su2/a)

    and another equal to

    exp(-iu - su2/a)

    can I complete the square with this and solve it?
     
  14. May 9, 2010 #13

    gabbagabbahey

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    Yes, start by completing the square...
     
  15. May 9, 2010 #14
    ok well..

    exp(iu - su2/a)

    = exp[-(su2/a - iu)]
    = exp[-(s/a)(u2 - i(a/s)u)]
    = exp[-(s/a)(u-(ia/2s)2 - (i2/(4(s/a))]
    = exp[-(s/a)(u-(ia/2s)2 + a /(4s)]
    ...?
     
  16. May 9, 2010 #15

    vela

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    Up to here it looks okay
    Your parentheses don't match up here, but what you actually have is probably right.

    Note that ea/(4s) is a constant, so you can pull it out of the integral.
     
  17. May 9, 2010 #16

    gabbagabbahey

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    Right, now [tex]e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2+\frac{a}{4s}}=e^{\frac{a}{4s}}e^{-\frac{s}{a}\left(u-\frac{ia}{2s}\right)^2}[/tex] and [tex]e^{\frac{a}{4s}}[/tex] is a constant and can be pulled out of the integral.

    Now, use the substitution [itex]v=\sqrt{\frac{s}{a}}\left(u-\frac{ia}{2s}\right)[/tex]....
     
  18. May 9, 2010 #17
    also.. I have a "u term" from what I did

    u = (at)1/2
    t = u2/a
    du = (a/2u)*dt
    dt = (2u/a)*du

    so I actually have a 1/ai in front of the integral and another u in it.
    That means I have to do integration by parts in addition to what you said, right?
     
  19. May 9, 2010 #18

    gabbagabbahey

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    You won't need to use IBP...what does your integral become after the v-sub?
     
  20. May 9, 2010 #19
    a big mess!!
    the limits of integration are from -(i/2)(a/s)1/2 to infinity now, and the equation under the integral is
    (a/s)1/2 * [(a/s)1/2v + (ia/2s)]exp(-v2) dv

    note: there was a u term before the whole exponential term, so I accounted for that. and then made a substitution for du as well
     
    Last edited: May 9, 2010
  21. May 9, 2010 #20

    gabbagabbahey

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    Well, I'm sure you know how to integrate [itex]\int ve^{-v^2}dv[/itex] right? And if you use the substitution [itex]v=\sqrt{\frac{s}{a}}\left(u+\frac{ia}{2s}\right)[/itex] on the [tex]e^{-\frac{s}{a}\left(u+\frac{ia}{2s}\right)^2[/tex]...I think you will get some fortuitous cancellation
     
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